UGC NET Computer Science Database Management Systems Previous Year Questions (PYQs)
Practise 37 Database Management Systems questions asked in UGC NET Computer Science from 2020–2025. Questions cover Data models, Relational algebra, Integrity constraints, Functional dependencies and normalisation, Transaction processing and Locking protocols. Every question is shown with its options and the correct answer, free to read.
37 PYQs2020–2025Answers includedFree
Practise Database Management Systems as timed sets
Focused 10-question sets with instant scoring, explanations, and weak-area analysis.
Which table contains the primary information in a data warehouse?
ADimension table
BFact table
CLookup table
DPrimary table
Answer:(B)Fact table
Explanation
A fact table stores the central measurable events and foreign keys in a warehouse schema. Dimension tables provide descriptive context for analysing those facts. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
A. Project-join normal form B. Lost-update problem C. Discretionary access control D. Data-manipulation language
I. Concurrency II. Fifth normal form III. SQL IV. Security
AA-II, B-I, C-IV, D-III
BA-I, B-II, C-IV, D-III
CA-I, B-II, C-III, D-IV
DA-IV, B-II, C-III, D-I
Answer:(A)A-II, B-I, C-IV, D-III
Explanation
Project-join normal form is fifth normal form; lost updates are a concurrency issue; discretionary access control is security; and SQL is a data-manipulation language. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Answer:(B)the number of attributes in the relation
Explanation
The degree (or arity) of a relation is its number of attributes/columns. The number of tuples/rows is called its cardinality. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
What is the correct sequence of phases of database design?
A. Physical design
B. Conceptual design
C. Logical design
D. Requirement collection and analysis
AD, A, B, C
BD, B, C, A
CD, B, A, C
DD, A, C, B
Answer:(B)D, B, C, A
Explanation
Database design starts with requirements, then conceptual modelling, logical schema design and finally physical design. This produces D, B, C, A. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
An alternate key in a database table is also called
APrimary key
BCandidate key
CSuper key
DForeign key
Answer:(B)Candidate key
Explanation
Candidate keys are minimal unique identifiers. One is selected as the primary key and the remaining candidate keys are alternate keys. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
In concurrency control, a phantom problem may occur when
Arecords are inserted
Brecords are deleted
Crecords are modified
Drecords are indexed
Answer:(A)records are inserted
Explanation
A phantom occurs when a transaction reruns a range query and sees new rows inserted by another transaction. Inserts change the membership of the queried set. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
A. add a single tuple to a relation
B. add multiple tuples to a relation
C. add values to specific attributes
D. insert a new table
AA only
BA and B only
CA, B and C only
DB, C and D only
Answer:(C)A, B and C only
Explanation
INSERT can add one row, multiple rows, or values for specified columns in a row. It does not create a table; CREATE TABLE is used for that purpose. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
A. A relation having only key attributes is always in BCNF, since it has no functional dependencies.
B. If a relation is not in 4NF because of a non-trivial multivalued dependency, it should be decomposed.
C. Decomposition can remove redundancy caused by a multivalued dependency.
D. Third normal form is stronger than BCNF.
AA only
BA and B only
CD only
DC only
Answer:(C)D only
Explanation
BCNF is stricter than 3NF, so statement D is false. Statements B and C correctly describe decomposition for non-trivial multivalued dependencies; statement A is true when all attributes are key attributes.
Which of the following commands can be used to modify data in an SQL table?
A. INSERT
B. DELETE
C. UPDATE
D. SELECT
AA, B and D only
BA, C and D only
CB, C and D only
DA, B and C only
Answer:(D)A, B and C only
Explanation
INSERT adds rows, DELETE removes rows and UPDATE changes rows. SELECT only reads data, so A, B and C are the modifying commands. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
A. Hadoop Distributed File System B. Veracity C. Apache Pig D. Apache Hive
I. Trust II. Facebook III. NameNode IV. Yahoo
AA-IV, B-III, C-I, D-II
BA-III, B-I, C-II, D-IV
CA-II, B-I, C-IV, D-II
DA-II, B-III, C-I, D-IV
Answer:(C)A-II, B-I, C-IV, D-II
Explanation
HDFS uses a NameNode, veracity is associated with trust, Apache Pig originated at Yahoo, and Apache Hive originated at Facebook. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
In a schema R(A, B, C, D, E, F, G, H), each field of R
contains only atomic values.
F= {CH→G, A →BC, B→CFH, E→A, F→EG} is a set
of functional dependencies F+ is exactly the set
of FDs that holds R. The relation R is :
AIn 1NF, but not in 2 NF
BIn 2 NF , but not in 3 NF
CIn 3NF, but not in BCNF
DIn BCNF
Answer:(A)In 1NF, but not in 2 NF
Explanation
D appears on no right-hand side, so every candidate key must contain it; taking closures gives AD, BD, ED and FD as the candidate keys. The prime attributes are therefore A, B, D, E and F, and the non-prime ones are C, G and H. But A → BC makes the non-prime attribute C depend on A alone, a proper subset of the key AD. That partial dependency breaks 2NF, while the relation is already in 1NF because every field is atomic. Option A is correct.
Consider a relation schema R=(U, V, W, X, Y, Z), on
which the following functional dependencies
hold
{U→V, VW→X, Y→W; X→U}
The candidate keys of R are :
AUY, VY
BUY, VY, XY
CUYZ, VYZ, VWZ
DUYZ, VYZ, XYZ
Answer:(D)UYZ, VYZ, XYZ
Explanation
Z appears on no right-hand side, so it must belong to every candidate key. Taking closures, UY⁺ gains V from U → V, W from Y → W and X from VW → X; VY⁺ and XY⁺ reach the same set. Each covers everything except Z, so appending Z gives the candidate keys UYZ, VYZ and XYZ. Option D is correct.
Bits are represented on the medium at the physical layer, so (A)-(II). Physical (MAC) addressing belongs to the data-link layer, giving (B)-(IV). Logical (IP) addressing belongs to the network layer, so (C)-(III). Segmentation and reassembly is a transport-layer duty, so (D)-(I). Option B matches all four.
Consider a schema R(P, Q, R, S) and the following
functional dependencies P → Q, Q → R, R → S, S
→ Q. Then decomposition of R (P, Q), R (Q, R)
and R 3(Q, S) is :
ADependency Preserving and lossless join
BLossless Join but not dependency preserving
CDependency preserving but not lossless Join
DNot dependency preserving and not lossless join
Answer:(A)Dependency Preserving and lossless join
Explanation
The sub-relations share Q, and Q determines R and, through R → S, also S, so rejoining them reconstructs R without spurious tuples: the decomposition is lossless. P → Q survives in R₁(P, Q), Q → R in R₂(Q, R), and R → S together with S → Q is recoverable from R₂ and R₃(Q, S). Every dependency remains enforceable, so the decomposition is both dependency preserving and lossless — option A.
What is SQL primarily used for in the context of
relational databases ?
ATo design user interfaces
BTo create and manipulate databases
CTo display data on web pages
DTo format printed reports
Answer:(B)To create and manipulate databases
Explanation
SQL is the standard language for defining and manipulating relational data: creating tables, inserting, updating and deleting rows, and querying with SELECT. User interfaces, page rendering and report formatting are the work of other tools. Option B is correct.
How does a relational database ensure data
integrity ?
ABy encrypting all data stored
BBy enforcing rules defined in the schema
CBy compressing data for efficient storage
DBy allowing unrestricted access to all users
Answer:(B)By enforcing rules defined in the schema
Explanation
Relational databases protect integrity through schema-level constraints — primary keys, foreign keys, unique, not-null and check constraints — which the engine enforces on every write. Encryption protects confidentiality, compression saves space, and unrestricted access would weaken integrity rather than ensure it. Option B is correct.
Fifth normal form, also called project-join normal form, is concerned with join dependency: a relation is in 5NF when it cannot be decomposed further without losing information. Multivalued dependency belongs to 4NF, and functional dependency to 2NF, 3NF and BCNF. Option A is correct.
Consider the transactions T₁, T₂, T₃, and the
schedules S₁ and S₂ given below.
T₁: r₁(x); r₁(z); w₁(z)
T₂: r₂(y); r₂(z); w₂(z)
T₃: r₃(y); r₃(x); w₃(y)
S₁ : r₁(x); r₃(y); r₃(x); r (y); r₂(z); w (y); w (z);
r (z); w (x); w (z)
S₂ : r₁(x); r₃(y); r (y), r₃(x); r₁(z); r₂(z); w₃(y);
w₁(x); w₂(z); w₁(z)
Which one of the following statements about the
schedules is TRUE?
AOnly S₁ is conflict-serializable
BOnly S₂ is conflict-serializable
CBoth S₁ and S₂ are conflict-serializable
DNeither S₁ nor S₂ is conflict-serializable
Answer:(A)Only S₁ is conflict-serializable
Explanation
Constructing the precedence graph for S₁ from its read-write, write-read and write-write conflicts produces no cycle, so S₁ is conflict-serializable. The graph for S₂ does contain a cycle, so no serial order is conflict-equivalent to it. Only S₁ qualifies — option A.
For R(V, W, X, Y, Z) with {VW→X, WX→V, VX→W, W→Y, Y→Z}, the closure of VW reaches every attribute, so VW is a candidate key. But W → Y and Y → Z both have non-superkey left-hand sides, violating BCNF. Any decomposition that removes them separates W from Y or Y from Z, and the lost dependency cannot be re-derived from the fragments. This relation therefore admits no lossless, dependency-preserving BCNF decomposition — option A.
Which of the following best describes the
structure of a relational database ?
AData organized into tables with rows and columns
BData organized into files and folders
CData organized into a hierarchical tree structure
DData organized into a network of interconnected nodes
Answer:(A)Data organized into tables with rows and columns
Explanation
A relational database stores data in tables of rows and columns, where a row is a record and a column an attribute, with relationships expressed through keys. Files and folders, tree hierarchies and node networks describe other data models. Option A is correct.
Paging maps logical addresses onto physical frames, so (A)-(III). LRU evicts the least recently used entry, giving (B)-(I). C-SCAN sweeps the disk in one circular direction, so (C)-(IV). Virtual memory makes the address space appear larger than physical memory, so (D)-(II). Option B is correct.
Arrange the following phases of database design
in the correct order :
A. Physical Design
B. Conceptual Design
C. Logical Design
D. Requirement Analysis
A(B), (D), (A), (C)
B(C), (A), (B), (D)
C(D), (B), (C), (A)
D(A), (D), (C), (B)
Answer:(C)(D), (B), (C), (A)
Explanation
Database design proceeds: requirement analysis (D), conceptual design (B) producing an ER model, logical design (C) mapping that model to relations, then physical design (A) choosing storage structures and indexes. That order is option C.
The key idea is stated in the problem: The Hue of a colour is related to its :. The correct answer is D: Wavelength. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.
Which of the statement are CORRECT ?
A. Constructors are invoked automatically when the objects are created.
B. Constructors do not have return types, not even void and therefore they cannot return values.
C. Constructors cannot be inherited though a derived class can call the base class constructors.
D. Constructors can be declared as virtual. Choose the correct answer from the options given below :
Which one of the following statements are CORRECT ?
A. Granularity is the size of data item in a database.
B. Two operations in a schedule are said to be conflict if they belong to same transaction.
C. Two schedulers are said to be conflict equivalent if the order of any two conflicting operations is the same in both schedules.
D. Write operations which are performed without performing the write operation are known as Blind Writes. Choose the correct answer from the options given below :
A(A) and @) Only
B(A), (B) and (C) Only
C(A), (B) and (D) Only
D(B) and (C) Only
Answer:(B)(A), (B) and (C) Only
Explanation
The key idea is stated in the problem: Which one of the following statements are CORRECT ?
A. Granularity is the size of data item in a database.
B. Two operations in a schedule are said to be conflict if they belong to same transaction.
C. Two schedulers are said to be conflict equivalent if th. The correct answer is B: (A), (B) and (C) Only. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.
Which of the following is/are NOT CORRECT statement ?
A. The first record in each block of the data file is known as actor record.
B. Dense index has index entries for every search key value in the data file.
C. Searching is harder in the B* tree than B— tree as the all external nodes linked to each other.
D. In extendible hashing the size of directory is just an array of 24—, where d is global depth. Choose the correct answer from the options given below :
The key idea is stated in the problem: In most general case, the computer needs to process each instruction with the following sequence of steps :
A. Calculate the effective address
B. Execute the instruction
C. Fetch the instruction from memory
D. Fetch the operand from memory
E. Decode the i. The correct answer is C: C, E, A, D, B. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.
Answer:(A)(D) Depth first search (IV) O(b' m) Choose the correct answer from the options given below :
Explanation
The key idea is stated in the problem: Match List - I with List - II. List - I List - II
A. Hillclimbing () Ob' d)
B. Best first search (I) O(bd)
C. A* Search (M1) O. The correct answer is A: (D) Depth first search (IV) O(b' m) Choose the correct answer from the options given below :. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.
For S = (U, V, W, X, Y, Z), the functional dependencies are {U → V, VW → X, Y → W, X → U}. Which listed sets are candidate keys?
AUY, VY
BUY, VY, XY
CUYZ, VYZ, VWZ
DUYZ, VYZ, XYZ
Answer:(D)UYZ, VYZ, XYZ
Explanation
Z never appears on the right-hand side, so every key must contain Z. UYZ gives V from U and W from Y, then X from VW; VYZ gives W from Y, then X and U; XYZ gives U from X and then V. Each is minimal, yielding option D.
Two operations from a schedule conflict when
A. at least one operation is a write,
B. both access the same data item, and
C. they belong to different transactions.
A(A) and (B) only
B(A), (B), and (C) only
C(A), (C), and (D) only
D(C) and (D) only
Answer:(B)(A), (B), and (C) only
Explanation
Conflicting operations are operations of different transactions on the same item where at least one writes. Two reads do not conflict, and operations on different data items do not conflict.
List I (E-R symbols) List II (Description)
A. () Key Attribute Type
— 33 (I) Weak Entity Type
<< > (II) Total Participation of Entity in a
C. relation
) <> E, (IV) Multivalue Attribute type
List I
List II
Items are listed in the question stem.
Match each item to its stated description or complexity.
AOCR review required
BA-IV, B-I, C-II, D-III
CA-II, B-I, C-IV, D-III
DA-III, B-IV, C-I, D-II
Answer:(C)A-II, B-I, C-IV, D-III
Explanation
The standard E–R notation distinguishes key, weak entity, total participation, and multivalued attributes by their respectively marked symbols; option C gives that correspondence.
Using the university table structures, which query returns distinct employee IDs of employees working on projects 20, 30, or 40?
ASELECT EMPID FROM PROJECTWORK WHERE PROJNO = (20,30,40);
BSELECT EMPID FROM PROJECTWORK WHERE PROJNO IN (20,30,40);
CSELECT DISTINCT EMPID FROM PROJECTWORK WHERE PROJNO IN (20,30,40);
DSELECT DISTINCT EMPID FROM PROJECTWORK WHERE PROJNO = 20,30,40;
Answer:(C)SELECT DISTINCT EMPID FROM PROJECTWORK WHERE PROJNO IN (20,30,40);
Explanation
IN tests membership in a set of values, and DISTINCT removes duplicate employee IDs when an employee works on more than one listed project. Equality cannot be used with a comma-separated list.
Which statement correctly computes SUM, MAX, MIN, and AVG salaries for the English department?
Statement I joins EMPLOYEE and DEPARTMENT on DEPTNO = DID and filters DNAME = 'ENGLISH'.
Statement II filters DNAME = 'ENGLISH' without joining the two tables.
ABoth statements are true.
BBoth statements are false.
CStatement I is true but Statement II is false.
DStatement I is false but Statement II is true.
Answer:(C)Statement I is true but Statement II is false.
Explanation
Statement I supplies the join predicate that relates an employee to a department before filtering by department name. Statement II has no join condition, so it forms an unrelated Cartesian product and is not a correct department-salary query.
Which query returns employee ID and name for employees whose salary is greater than the salary of every employee in department 20, ordered by employee ID?
A. uses SALARY > (subquery) with ORDER BY.
B. uses SALARY > (subquery) without ORDER BY.
C. uses SALARY > ALL (subquery) with ORDER BY.
A(A) and (B) only
B(A) and (C) only
C(B) only
D(C) only
Answer:(D)(C) only
Explanation
A scalar comparison with a subquery returning several salaries is not the required “greater than every” test and can be invalid. > ALL (...) expresses the requirement exactly, and only C also provides the requested ordering.
Which SQL statement(s) can drop the SALARY column from EMPLOYEE?
A. ALTER TABLE EMPLOYEE DROP SALARY CASCADE;
B. ALTER TABLE EMPLOYEE DROP SALARY RESTRICT;
C. ALTER EMPLOYEE DROP SALARY;
A(A) and (B) only
B(A) and (C) only
C(B) and (C) only
D(A) only
Answer:(A)(A) and (B) only
Explanation
The supported form is ALTER TABLE ... DROP .... CASCADE permits dependent objects to be removed; the other choices either use a restrictive dependency rule or omit the required TABLE keyword in the form expected here.
Using the university table structures, the university wants to give every employee in the SCIENCE department a 20% salary rise. Which query is correct?
A. UPDATE EMPLOYEE SET SALARY = SALARY * 1.20 WHERE DEPTNO IN (SELECT DID FROM DEPARTMENT WHERE DNAME = 'SCIENCE');
B. UPDATE TABLE EMPLOYEE SET SALARY = SALARY * 1.20 WHERE DNAME = 'SCIENCE';
C. ALTER TABLE EMPLOYEE SET SALARY = SALARY * 1.20 WHERE DEPTNO IN (SELECT DNAME FROM DEPARTMENT WHERE DNAME = 'SCIENCE');
A(A) and (B) only
B(A) only
C(B) and (C) only
D(C) only
Answer:(B)(A) only
Explanation
A is valid: it updates EMPLOYEE rows whose department number belongs to the SCIENCE department. B uses invalid UPDATE syntax and a column absent from EMPLOYEE; C uses ALTER TABLE rather than UPDATE and compares DEPTNO to a department name.
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Frequently asked questions
How many Database Management Systems questions have been asked in UGC NET Computer Science?
37 Database Management Systems questions appear in the UGC NET Computer Science papers held between 2020–2025, and all of them are on this page with their answer key.
Are the answers on this page free?
Yes. Every question, its options, and the correct answer are free to read with no account. Signing in additionally unlocks the detailed explanation under each question.
Is Database Management Systems an important topic for UGC NET Computer Science?
Database Management Systems appears in every recent UGC NET Computer Science paper, across all 4 sittings covered here. Its share of the paper makes it worth revising thoroughly rather than sampling.
How should I practise Database Management Systems after reading these questions?
Attempt the Database Management Systems topic-wise sets, which put the same questions into a timed interface with instant scoring and weak-area analysis afterwards.